Check if the sequence is convergent

Question

Check if the sequence $$\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}$$ is convergent. I really don't know how to start.

Answer 1

It is decreasing and bounded! You can check it yourself.

Answer 2

Draw a graph, and notice that

$$\sum_{k=n}^{2n}\frac1k\le\int_{n-1}^{2n}\frac{dx}x=\ln 2n-\ln(n-1)=\ln\frac{2n}{n-1}=\ln\left(2+\frac2{n-1}\right)\;.$$

On the other hand,

$$\sum_{k=n}^{2n}\frac1k\ge\int_n^{2n+1}\frac{dx}x=\ln(2n+1)-\ln n=\ln\left(2+\frac1n\right)\;.$$

Answer 3

These some methods:

• Use this result

$$H_n=\sum_{k=1}^n \frac1k=\ln n+\gamma+o(1)$$ and notice that the given sum is $H_{2n}-H_{n-1}$.

• Write the given sum on the Riemann sum
• Notice that the given sum is a monotonic sequence bounded by $1$ and $\frac12$.

The two first method give also the limit: $\ell=\ln 2$ and the third method gives this estimation of the limit

$$\frac12\le\ell\le1$$

Answer 4

$$\frac{1}{n+n}\leq \frac{1}{n}\leq \frac{1}{n}\\\frac{1}{n+n}\leq \frac{1}{n+1}\leq \frac{1}{n}\\\frac{1}{n+n}\leq \frac{1}{n+2}\leq \frac{1}{n}\\\frac{1}{n+n}\leq \frac{1}{n+3}\leq \frac{1}{n}\\.\\.\\.\\\frac{1}{n+n}\leq \frac{1}{n+n}\leq \frac{1}{n}\\\\sumation\\n\frac{1}{n+n}\leq \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...\frac{1}{n+n}\leq n\frac{1}{n}\\\frac{n}{n+n}\leq \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...\frac{1}{n+n}\leq \frac{n}{n}\\\frac{1}{2}\leq \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...\frac{1}{n+n}\leq \frac{1}{1}\\ $$ $$\lim_{n\rightarrow \infty }a_{n}=\lim_{n\rightarrow \infty }\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}=\\=\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\frac{1}{1+\frac{3}{n}}+...+\frac{1}{1+\frac{n}{n}})=\\\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+\frac{i}{n}}=\\\int_{0}^{1}\frac{1}{1+x}dx=ln(1+x) \\=ln(2)$$

Answer 5

Let our sum be $a_n$. It is clear that the $a_n$ are bounded below by $0$.

We show the sequence $(a_n)$ is decreasing. Note that $$a_{n+1}-a_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n}\lt \frac{2}{2n+1}-\frac{2}{2n}\lt 0.$$

Answer 6

$$\lim_{n\rightarrow \infty }a_{n}=\lim_{n\rightarrow \infty }\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}=\\=\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\frac{1}{1+\frac{3}{n}}+...+\frac{1}{1+\frac{n}{n}})=\\\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+\frac{i}{n}}=\\\int_{0}^{1}\frac{1}{1+x}dx=ln(1+x) \\=ln(2)$$